Let $F$ be a field and $\phi: F[x] \to F[x]$ be a ring (with identity) isomorphism.
Suppose $\phi(a) \in F$ for all $a \in F$, where $F$ represents the constant polynomials. I want to show that the degree of $\phi(x) = 1$.
To begin, let $k$ be the degree of $\phi$, and let $f(x) = a_0 + ... + a_n x^n \in F[x]$ be an element such that $\phi(f(x)) = x$ (with $a_k$ nonzero). Next, I know I need to evaluate $\phi$ at $a_0 +...+ a_nx^n$ to compute the degree of $\phi(f(x))$, then show $n=k=1$. I am unsure of how to explicitly do this.
Just take $f(x) = a_0 + \cdots + a_n x^n$. Notice that
$$\phi(f(x)) = \phi(a_0) + \cdots + \phi(a_n) \phi(x)^n$$
using the fact that it's a ring isomorphism. Now if the degree of $\phi(x)$ is not $1$, then I claim this breaks surjectivity. For example, if $\phi(x) \in F$ (so degree $0$) then this maps to $F$ and we fail surjectivity. If $\phi(x) = 0$ (so degree $-\infty$ depending on your context), then again surjectivity breaks. If $\deg(\phi) > 2$, then we see that degree $1$ polynomials are missed entirely by the above calculation, so surjectivity breaks. That is, since $\phi(a_n) \neq 0$ (as we assume $a_n \neq 0$) we have that $\deg(\phi(f(x)) = \deg(\phi(x)) \cdot \deg(f).$